can u help me out understanding the exit case for this recursion program??

Question

def primecheck(a,i = 2):
if a <= 1:
return("Nonprime!!")
elif a == 2:
return("Prime")
elif (a % i == 0):
return("nonprime")
elif (i*i > a):
return("prime")
return primecheck(a,i+1)

#main..

num = int(input("Enter the number: "))
print(primecheck(num))

in this code can anyone make me understand how the recursion loop ends and output comes the entered number i >2 and prime??

Answer 1

The function primecheck(), given a number a, will check for every possible divisor i, inside the range 2 (the default second argument) and i*i (no need to go further on), incrementing i by 1 at every new recursion (can be optimized changing the increment: 1 is a lot of recursions).

Before recursing, there are 4 exit cases:
- the given number is 0 or 1: it is said not prime
- the given number is 2: prime
- found a divisor i: prime
- reached the end of possible divisors range i*i, without having found any divisor: prime

Answer 2

Thanks a lot got that